C Program to Convert Binary to Octal

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In this tutorial, we will learn about how to create a program in C that can convert any given binary number (by user at run-time) into its equivalent octal number. At last we have also created a function named BinToOct() that does the same job.

But before going through the program, if you are not aware of

Binary Number
Octal Number
Binary to Octal Conversion

then refer to Binary to Octal conversion step by step process. Now let's move on to the program.

Binary to Octal in C

The question is, Write a program in C that converts any given binary number into its equivalent octal value. The answer to this question is:

// Binary to Octal in C
// ----codescracker.com----

#include<stdio.h>
#include<conio.h>
int main()
{
    int bin, oct=0, i=0, mul=1, count=1, rem, octnum[20];
    printf("Enter any Binary Number: ");
    scanf("%d", &bin);
    while(bin!=0)
    {
        rem = bin%10;
        oct = oct + (rem*mul);
        if(count%3==0)
        {
            octnum[i] = oct;
            mul = 1;
            oct = 0;
            count = 1;
            i++;
        }
        else
        {
            mul = mul*2;
            count++;
        }
        bin = bin/10;
    }
    if(count!=1)
        octnum[i] = oct;
    printf("\nEquivalent Octal Value = ");
    for(i=i; i>=0; i--)
        printf("%d", octnum[i]);
    getch();
    return 0;
}

As the above program was written under Code::Blocks IDE, therefore after successful build and run, here is the first snapshot of the sample run:

Now supply any binary number say 1101110 as input and press ENTER key to see its equivalent octal value as shown in the second snapshot of the sample run given below:

Here is the final snapshot of another sample run:

Program Explained

Receive any binary number from user at run-time say 1101110
Create a while loop that runs until the value of binary number (bin) becomes 0
At first run, bin (1101110) is not equal 0, therefore program flow goes inside the loop
And bin%10 or 1101110%10 or 0 gets initialized to rem, oct + (rem*mul) (we have initialized 0 to oct and 1 to mul at start of the program) or 0 + (0*1) or 0 gets initialized to oct
We have initialized 1 to count at start of the program, therefore at first run, the value of count is not equal to 3, therefore while dividing the value of count or 1 with 3, we will not get a remainder as 0, therefore program flow goes inside the else block (as the condition of if statement evaluates to be false at first run of while loop), and inside the else block, mul*2 or 1*2 or 2 gets initialized to mul, and the value of count gets incremented and becomes 2
At last, bin/10 or 1101110/10 or 110111 gets initialized to bin, and again program flow goes back to the while loop's condition
Therefore, bin!=0 or 110111!=0 evaluates to be true, therefore program flow again goes inside the loop
And bin%10 or 110111%10 or 1 gets initialized to rem, and oct + (rem*mul) or 0 + (1*2) or 2 gets initialized to oct
Again the condition of if statement, that is count%3==0 or 2%3==0 evaluates to be false, therefore program flow goes inside the else block, and mul*2 or 2*2 or 4 gets initialized to mul, and the value of count gets incremented and becomes 3
At last, bin/10 or 110111/10 or 11011 gets initialized to bin, and again program flow goes back to the loop's condition
Therefore, bin!=0 or 11011!=0 evaluates to be true, therefore program flow again goes inside the loop
And bin%10 or 11011%10 or 1 gets initialized to rem, and oct + (rem*mul) or 2 + (1*4) or 6 gets initialized to oct
Now the condition of if statement, that is count%3==0 or 3%3==0 evaluates to be true, therefore program flow goes inside the if block, and oct or 6 gets initialized to octnum[i] (we have initialized 0 to i at start of the program) or octnum[0], and 1, 0, 1 gets initialized to mul, oct, count respectively. And the value of i gets incremented and becomes 1
At last, bin/10 or 11011/10 or 1101 gets initialized to bin, and again program flow goes back to the loop's condition
Therefore, bin!=0 or 1101!=0 evaluates to be true, therefore program flow again goes inside the loop and follow the same procedure as told above
After exiting from the while loop, check whether the value of count is not equal to 1 or not, if it is, then the value of oct gets initialized to the last index of octnum[] array
Finally print the value of of octal digit one by one from last index to first index

using User-defined Function

The question is, Write a program in C that converts Binary to Octal using user-defined function named BinToOct(). Here we have declared the variable i and the array octnum[20] as global variable (out from both the function) to make it known inside both the function that is main() and BinToOct() function.

// Binary to Octal in C using Function
// ----codescracker.com----

#include<stdio.h>
#include<conio.h>
void BinToOct(int bin);
int i=0;
int octnum[20];
int main()
{
    int binnum;
    printf("Enter any Binary Number: ");
    scanf("%d", &binnum);
    BinToOct(binnum);
    printf("\nEquivalent Octal Value = ");
    for(i=i; i>=0; i--)
        printf("%d", octnum[i]);
    getch();
    return 0;
}
void BinToOct(int bin)
{
    int oct=0, mul=1, count=1, rem;
    while(bin!=0)
    {
        rem = bin%10;
        oct = oct + (rem*mul);
        if(count%3==0)
        {
            octnum[i] = oct;
            mul = 1;
            oct = 0;
            count = 1;
            i++;
        }
        else
        {
            mul = mul*2;
            count++;
        }
        bin = bin/10;
    }
    if(count!=1)
        octnum[i] = oct;
}

Here is the final snapshot of above program's sample run:

Same Program in Other Languages

C Online Test

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